Solving F(x) = X^2 - 4x + 2 And Finding F(a) = 2

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Hey guys! Today, we're diving deep into the fascinating world of quadratic functions, specifically focusing on the function f(x) = x^2 - 4x + 2. We'll not only explore the function itself but also tackle the problem of finding the value of 'a' when f(a) = 2. Get ready to sharpen your pencils and flex those brain muscles – let's get started!

Understanding Quadratic Functions

Before we jump into the specifics, let's establish a solid foundation by understanding what quadratic functions are all about. A quadratic function is a polynomial function of the second degree, meaning the highest power of the variable (in our case, 'x') is 2. The general form of a quadratic function is f(x) = ax^2 + bx + c, where 'a', 'b', and 'c' are constants, and 'a' is not equal to 0. These constants play crucial roles in determining the shape and position of the parabola, which is the graph of a quadratic function.

Key characteristics of quadratic functions include:

  • Parabola: The graph of a quadratic function is a U-shaped curve called a parabola. This parabola can open upwards (if 'a' is positive) or downwards (if 'a' is negative).
  • Vertex: The vertex is the highest or lowest point on the parabola, depending on whether it opens upwards or downwards. It's a critical point for understanding the function's behavior.
  • Axis of Symmetry: This is a vertical line that passes through the vertex and divides the parabola into two symmetrical halves. It's like a mirror reflecting one side onto the other.
  • Roots/Zeros: These are the x-values where the parabola intersects the x-axis. They are also known as the solutions to the quadratic equation f(x) = 0.
  • Y-intercept: This is the point where the parabola intersects the y-axis. It's found by setting x = 0 in the function.

In our specific function, f(x) = x^2 - 4x + 2, we have a = 1, b = -4, and c = 2. This means our parabola will open upwards since 'a' is positive. Now that we have a good grasp of the fundamentals, let's delve deeper into analyzing this particular quadratic function.

Analyzing f(x) = x^2 - 4x + 2

To truly understand f(x) = x^2 - 4x + 2, we need to analyze its key features. We'll start by finding the vertex, axis of symmetry, and y-intercept. These elements will give us a clear picture of the parabola's shape and position in the coordinate plane. Remember, understanding these features is crucial for solving various problems related to quadratic functions, including the one we'll tackle later: finding 'a' when f(a) = 2.

Finding the Vertex

The vertex is arguably the most important point on a parabola. It represents the minimum or maximum value of the function. The x-coordinate of the vertex can be found using the formula x = -b / 2a. In our case, a = 1 and b = -4, so the x-coordinate of the vertex is x = -(-4) / (2 * 1) = 2. To find the y-coordinate, we substitute this x-value back into the function: f(2) = (2)^2 - 4(2) + 2 = 4 - 8 + 2 = -2. Therefore, the vertex of our parabola is at the point (2, -2).

Determining the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two symmetrical halves. Its equation is simply x = (the x-coordinate of the vertex). So, for our function, the axis of symmetry is the line x = 2. This line acts like a mirror, reflecting the parabola across itself.

Locating the Y-intercept

The y-intercept is the point where the parabola crosses the y-axis. To find it, we set x = 0 in the function: f(0) = (0)^2 - 4(0) + 2 = 2. Therefore, the y-intercept is at the point (0, 2). This point gives us another reference point for sketching the graph of the parabola.

Finding the Roots (x-intercepts)

The roots, or x-intercepts, are the points where the parabola intersects the x-axis. To find them, we need to solve the quadratic equation f(x) = 0. In other words, we need to find the values of 'x' that make the function equal to zero. There are several ways to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. For our function, f(x) = x^2 - 4x + 2, factoring is not straightforward, so let's use the quadratic formula.

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a. Plugging in our values a = 1, b = -4, and c = 2, we get:

x = (4 ± √((-4)^2 - 4 * 1 * 2)) / (2 * 1) x = (4 ± √(16 - 8)) / 2 x = (4 ± √8) / 2 x = (4 ± 2√2) / 2 x = 2 ± √2

So, the roots of our function are x = 2 + √2 and x = 2 - √2. These are the points where the parabola crosses the x-axis. Knowing the roots, vertex, axis of symmetry, and y-intercept gives us a comprehensive understanding of the graph of f(x) = x^2 - 4x + 2.

Solving f(a) = 2

Now, let's tackle the main problem: finding the value(s) of 'a' such that f(a) = 2. This means we need to substitute 'a' for 'x' in our function and set the result equal to 2. We'll then solve the resulting equation for 'a'. This is a classic application of understanding function evaluation and solving quadratic equations.

We have f(a) = a^2 - 4a + 2. Setting this equal to 2, we get:

a^2 - 4a + 2 = 2

Subtracting 2 from both sides, we get:

a^2 - 4a = 0

Now, we can factor out an 'a' from the left side:

a(a - 4) = 0

For this equation to be true, either a = 0 or a - 4 = 0. Solving for 'a' in the second equation, we get a = 4. Therefore, the values of 'a' that satisfy f(a) = 2 are a = 0 and a = 4. This means that when we input 0 or 4 into the function, the output will be 2. We've successfully found the solutions!

Visualizing the Solution

To solidify our understanding, let's think about this graphically. We're essentially looking for the points on the parabola f(x) = x^2 - 4x + 2 where the y-coordinate is 2. We already know the y-intercept is (0, 2), which corresponds to our solution a = 0. The other solution, a = 4, represents another point on the parabola with a y-coordinate of 2. If you were to sketch the graph of the parabola and draw a horizontal line at y = 2, you'd see that the line intersects the parabola at two points: x = 0 and x = 4. This visual representation reinforces our algebraic solution.

Real-World Applications of Quadratic Functions

Quadratic functions aren't just abstract mathematical concepts; they have numerous real-world applications. Understanding them is crucial in various fields, from physics and engineering to economics and computer science. Let's explore some examples:

  • Projectile Motion: The path of a projectile (like a ball thrown in the air) can be modeled using a quadratic function. The function can help determine the projectile's maximum height, range, and time of flight. Engineers and physicists use these principles to design everything from rockets to sports equipment.
  • Optimization Problems: Quadratic functions are often used to solve optimization problems, where the goal is to find the maximum or minimum value of a quantity. For example, a business might use a quadratic function to model profit and determine the optimal price to charge for a product to maximize profit. Similarly, farmers can use quadratic functions to optimize crop yield based on factors like fertilizer input.
  • Bridge and Arch Design: The curves of bridges and arches are often parabolic, which means quadratic functions play a crucial role in their design. Engineers use these functions to ensure the structural integrity and stability of these structures.
  • Computer Graphics: Quadratic functions are used in computer graphics to create smooth curves and surfaces. They are essential for rendering realistic images and animations.
  • Economics: Quadratic functions can be used to model cost, revenue, and profit functions in economics. Understanding these relationships can help businesses make informed decisions about pricing, production, and investment.

These are just a few examples of the many applications of quadratic functions. By mastering these functions, you're not just learning math; you're gaining valuable tools for solving real-world problems.

Conclusion

We've journeyed through the intricacies of the quadratic function f(x) = x^2 - 4x + 2, from understanding its fundamental characteristics to solving the problem of finding 'a' when f(a) = 2. We've also touched upon the diverse real-world applications of quadratic functions. I hope this exploration has demystified these functions and empowered you to tackle similar problems with confidence. Remember, practice makes perfect, so keep exploring and experimenting with quadratic functions – you'll be amazed at what you can achieve! Keep up the great work, guys!